Basic Laplace Transform of a 1st order Differential Equation

Just wondering if someone could have a quick look over what I've done so far:

**Assuming zero initial conditions, solve the following DE:**

$\displaystyle \dot{x} + x = 2$

The Laplace Transform gives:

$\displaystyle sX(s) - x_0 + \frac{1}{s^2} = \frac{2}{s}$

Rearranging and using the fact that $\displaystyle x_0 =0$ gives us

$\displaystyle X(s) = \frac{2}{s^2} - \frac{1}{s^3}$

$\displaystyle X(s) = 2\frac{1!}{s^{1+1}} - \frac{1}{2} \frac{2!}{s^{2+1}}$

So $\displaystyle x(s) = 2t - \frac{t^2}{2}$?

Re: Basic Laplace Transform of a 1st order Differential Equation

After taking the Laplace transform, where did the $\displaystyle \frac{1}{s^2}$ come from?

Re: Basic Laplace Transform of a 1st order Differential Equation

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**Danny** After taking the Laplace transform, where did the $\displaystyle \frac{1}{s^2}$ come from?

I took the Laplace transform of $\displaystyle x$, is this not right?

Re: Basic Laplace Transform of a 1st order Differential Equation

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**craig** I took the Laplace transform of $\displaystyle x$, is this not right?

$\displaystyle x$ is a dependent variable ( $\displaystyle x=x(t)$ ), so you'll obtain

$\displaystyle s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots$

Re: Basic Laplace Transform of a 1st order Differential Equation

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**FernandoRevilla** $\displaystyle x$ is a dependent variable ( $\displaystyle x=x(t)$ ), so you'll obtain

$\displaystyle s\mathcal{L}\{x\}-x_0+\mathcal{L}\{x\}=\frac{2}{s}\Rightarrow (s+1)\mathcal{L}\{x\}=x_0+\frac{2}{s}\Rightarrow \ldots$

Thanks for the reply!

So given that $\displaystyle x_0 = 0$, does that mean that we just have $\displaystyle \mathcal{L}\{x\}=\frac{2}{s(s+1)}$.

Which I presume we express in partial fractions, and then solve for $\displaystyle x(t)$?

Re: Basic Laplace Transform of a 1st order Differential Equation

So for $\displaystyle \mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}$ we have:

$\displaystyle \mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}$

So $\displaystyle x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Thanks again for the replies.

Re: Basic Laplace Transform of a 1st order Differential Equation

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**craig** So $\displaystyle x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Right.

Re: Basic Laplace Transform of a 1st order Differential Equation

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**craig** So for $\displaystyle \mathcal{L}\{x\}=\frac{2}{s(s+1)} = \frac{2}{s} - \frac{2}{s+1}$ we have:

$\displaystyle \mathcal{L}\{x\}= 2\frac{1}{s} - 2\frac{1}{s-(-1)}$

So $\displaystyle x(t) = 2(1) - 2 (e^{-1t}) = 2(1-e^{-t)$ ?

Thanks again for the replies.

Technically, since you're working with the one-sided LT (I assume), you have unit step functions multiplying everything. That is,

$\displaystyle x(t)=2u(t)(1-e^{-t}),$

and the solution is valid for non-negative t's.

Re: Basic Laplace Transform of a 1st order Differential Equation

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**Ackbeet** and the solution is valid for non-negative t's.

Also considering the theorem about the necessary form of the solutions for the equation $\displaystyle x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t)$ .

Re: Basic Laplace Transform of a 1st order Differential Equation

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**FernandoRevilla** Also considering the theorem about the necessary form of the solutions for the equation $\displaystyle x^{(n)}+a_{n-1}x^{(n-1)}+\ldots +a_1x'+a_0x=b(t)$ .

Sorry would you mind explaining what you meant by this? I understood the first part about the Unit-Step function, just not sure what you meant here?

Thankyou again!

Re: Basic Laplace Transform of a 1st order Differential Equation

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**craig** Sorry would you mind explaining what you meant by this?

The Laplace transform of $\displaystyle x:[0,+\infty)\to\mathbb{R}$ is defined by $\displaystyle \mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt$ , so *a priori* we find solutions valid for $\displaystyle t\geq 0$. Your equation has (by a well known theorem) a determined form for its unique solution $\displaystyle x(t)=k+Ce^{at}$ valid for all $\displaystyle t\in\mathbb{R}$ so, *necessarily* is valid the solution found by LT for all $\displaystyle t\in\mathbb{R}$ .

Re: Basic Laplace Transform of a 1st order Differential Equation

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**FernandoRevilla** The Laplace transform of $\displaystyle x:[0,+\infty)\to\mathbb{R}$ is defined by $\displaystyle \mathcal{L}\{x(t)\}=\int_0^{+\infty}e^{-st}x(t)\;dt$ , so *a priori* we find solutions valid for $\displaystyle t\geq 0$. Your equation has (by a well known theorem) a determined form for its unique solution $\displaystyle x(t)=k+Ce^{at}$ valid for all $\displaystyle t\in\mathbb{R}$ so, *necessarily* is valid the solution found by LT for all $\displaystyle t\in\mathbb{R}$ .

Agreed. You could simply say that you extend the solution found by the LT method to the entire real line. The result of the inverse LT does have the unit step functions in it, and, in fact, does not satisfy the DE for negative t's. However, with the theorem you have invoked, you can extend the solution by eliminating the unit step function.

Re: Basic Laplace Transform of a 1st order Differential Equation

Thankyou both! I don't think we've covered this in lectures yet but I think it makes sense.